QQuestionPhysics
QuestionPhysics
3. Figure 2 shows a double-pane glass window. Double-pane glass windows are typically used to reduce heat loss from the inside of a structure to the outside. This is because a thin layer of *quiescent* air is trapped between both panes of glass.
**Figure 2.** Double-pane glass window (material is "boda-lime glass")
Assuming one-dimensional, steady-state conditions with constant properties, and using the information from **Figure 2 **
- a) Draw the thermal circuit that's representative of the heat transfer through the double-pane window.
- b) Determine the rate of heat flux *q'*<sup>H</sup> through the double-pane window (α<sup>H</sup> = 1).
*Hint to this problem – "quiescent" means stagnant (motionless). The trapped air isn't moving. Recall that convection = advection + diffusion. No fluid motion means no advection. So do NOT analyze the trapped air as convection.*
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Answer
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Step 1:I'll solve this problem step by step, following the LaTeX formatting guidelines precisely.
Step 2:: Analyze the Double-Pane Window Components
- Temperature difference: $$\Delta T = 40 \mathrm{~K}
The window consists of: - Outer glass pane (boda-lime glass) - Air gap (quiescent/stagnant) - Inner glass pane (boda-lime glass) Given information:
Step 3:: Draw Thermal Circuit
- $$A$$ is cross-sectional area
The thermal circuit for this scenario will have three thermal resistances in series:
Step 4:: Calculate Thermal Resistances
R_{3} = \frac{t_{3}}{k_{glass} \cdot A} = \frac{4 \times 10^{-3}}{1 \cdot 1} = 4 \times 10^{-3} \mathrm{~K/W}
a) Outer glass pane resistance: b) Air gap resistance: c) Inner glass pane resistance:
Step 5:: Calculate Total Thermal Resistance
R_{total} = R_{1} + R_{2} + R_{3} = (4 \times 10^{-3}) + 0.462 + (4 \times 10^{-3}) = 0.470 \mathrm{~K/W}
Step 6:: Calculate Heat Flux
q'^{H} = \frac{40}{0.470} = 85.11 \mathrm{~W/m^{2}}
Final Answer
The rate of heat flux q'^{H} through the double-pane window is 85.11 \mathrm{~W/m^{2}}. Key Insights: - The air gap provides significant thermal resistance - Quiescent air has low thermal conductivity - Thermal resistance is calculated by thickness divided by thermal conductivity
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